17.2 Relation Between Mean Power and Maximum Power for an Alternating Current

17.2 Relation Between Mean Power and Maximum Power for an Alternating Current

An alternating current is a current that periodically reverses direction, causing the voltage and current to fluctuate sinusoidally. A resistive load is one that does not store energy (like a capacitor or inductor) and simply dissipates energy as heat.

The instantaneous power $P$ in a circuit is the power at any given moment in time $t$.

where $V$ is the instantaneous voltage and $I$ is the instantaneous current. The maximum or peak $P_{\max}$ occurs when the current and voltage are at their peak values, $I_{o}$ and $V_{o}$ respectively. Also the mean power $P_{\mathrm{mean}}$ is the average power over one complete cycle of the AC waveform as shown in Fig. 17.4.

(Fig. 17.4: Mean power is exactly half the maximum power)

For a sinusoidal AC, let $I_{o}$ be the maximum or peak value of the current while voltage $V_{o}$ is the maximum or peak value of the voltage, then the maximum or peak power $P_{\max}$ is given by

Also we have expression for the mean power $P_{\mathrm{mean}}$ represented by

As $I_{o} = \sqrt{2}\, I_{\mathrm{rms}}$, so, using Eq. 17.6 becomes: $P_{\max} = (\sqrt{2} I_{\mathrm{rms}})^{2} R = 2 I_{\mathrm{rms}}^{2} R$ From Eq. 17.7 ; $I R_{\mathrm{rms}}^{2} = P_{\mathrm{mean}}$ , thus $P_{\max} = 2 P_{\mathrm{mean}}$ or

Therefore, it can be concluded that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current or voltage because the instantaneous power in an AC circuit varies over time. While the instantaneous power reaches at maximum value at certain points in the cycle, it spends significant time at zero or lower values, resulting in an average power that is half the peak value.

Interesting Information

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Digital Clamp Meter For Measuring AC and DC

Example 17.2

An alternating voltage supplied across a resistor of $50\,\Omega$ has a voltage of $220\,\mathrm{V}$. Calculate the mean power in kilo watt of the system.

Solution

$R = 50\,\Omega$ and $V = 220\,\mathrm{V}$ We know that; $P_{\max} = \dfrac{V^{2}}{R}$ Putting the values $P_{\max} = \dfrac{(220\,\mathrm{V})^{2}}{50\,\Omega} = 968\,\mathrm{W}$ As the mean power is half of the maximum (peak) power, $P_{\mathrm{mean}} = \dfrac{P_{\max}}{2} = \dfrac{968\,\mathrm{W}}{2}$ $= 484\,\mathrm{W}$ or $0.484\,\mathrm{kW}$

Fascinating Fact

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Scientists have converted human blood sugar into electricity